If f(x)=6arcsin(x^3) find f'(x)
Use the rule for the deriviative of a function of a function.
Let u(v) = 6 arcsinv and v(x)= x^3
f(x) = u(v(x))
f'(x) = du/dv * dv/dx
= [6/sqrt(1-v^2)]* 2x^2
= 12 x^2/sqrt(1-x^6)]
18 x^2/sqrt(1-x^6)
You're right; I typed the wrong coefficient for dv/dx
To find the derivative of f(x) = 6arcsin(x^3), we can apply the chain rule, which states that if we have a composition of functions, the derivative is the derivative of the outer function times the derivative of the inner function.
Let's break down the function into the outer function and the inner function:
Outer function: f(x) = 6arcsin(u)
Inner function: u = x^3
Now, let's find the derivative of each part:
Step 1: Find the derivative of the inner function u = x^3:
Using the power rule, we can differentiate the function with respect to x:
du/dx = d/dx(x^3) = 3x^2
Step 2: Find the derivative of the outer function f(u) = 6arcsin(u):
To differentiate the arcsin function, we use the chain rule again. The derivative of arcsin(u) with respect to u is 1/sqrt(1-u^2). Therefore, we have:
df/du = 1/sqrt(1-u^2)
Step 3: Apply the chain rule:
To find f'(x), we multiply the derivatives of the outer and inner functions:
f'(x) = (df/du) * (du/dx)
= (1/sqrt(1-u^2)) * (3x^2)
Finally, substitute back the value of u:
f'(x) = (1/sqrt(1-(x^3)^2)) * (3x^2)
Therefore, the derivative of f(x) = 6arcsin(x^3) is f'(x) = (1/sqrt(1-(x^3)^2)) * (3x^2).